Nice sir

@PreMath

Ай бұрын

Glad to hear that! Thanks ❤️

The use of proportions and complimentary angles. Thank you!

@PreMath

Ай бұрын

Yes! Thanks ❤️

Thanks for the question sir 😊

FC is 40% of BC; if BC=x we have (0.4x)^2 + x^2 = 20^2; 0.16x^2 + x^2 = 400; 1,16x^2=400; x^2=400/1.16=344,83 = area of blue square

@PreMath

Ай бұрын

Excellent! Thanks for sharing ❤️

Nice

Alternatively, length BC could be labelled as s. Then, length DF becomes (3/5)s, leaving (2/5)s for length CF. Applying Pythagoras to ΔBCF, s² + ((2/5)s)² = 20², which expands to s² + (4/25)s² = 400 and simplifies to (29/25)s² = 400 and area = s² = 10000/29 cm², as PreMath also found. In the video, the square of variable k is found and an additional step is required, multiplying by 25 to find the area of the square. One other thought: while k is a perfectly good choice for designating a variable, k is also very commonly used to designate 1000. So, in some viewers' minds, 3k may mean 3000 and they need to understand that k does not mean 1000 in this context. A different choice of letter should avoid this confusion.

@PreMath

Ай бұрын

Excellent! Thanks for sharing ❤️

AB=BC=a. 2*120/20=EF=12→ Razón de semejanza entre EDF y FCB =s=12/20=3/5→ DF=3a/5→ FC=2a/5→ a²+(2a/5)²=20²→ a²=10000/29. Gracias y saludos.

@PreMath

Ай бұрын

Excellent! You are very welcome! Thanks for sharing ❤️

Start at the point where EF is found to be 12 and the two triangles are found to be similar. From this point, I find the introduction of "K" makes the problem seem more complicated (perhaps this is a mental block of mine). I'll start with your equation DF/BC = 3/5. BC is the square side length, I'll call it "s". The other side of the BFC triangle is FC. I'll call this "x" and try to solve for x. With s and x defined, DF = s-x. So (s-x)/s = 3/5. Cross multiplying, 5(s-x) = 3s. 5s - 5x = 3s. Rearranging, 5x = 2s or x = 2s/5. That can be used in the pythagorean formula for BFC. s^2 + 4s^2/25 = 400. Or 29/25 * s^2 = 400. s^2 = 400*25/29 or s^2 = 10000/29 as you found. Is it easier using K? Not for me anyway.

@PreMath

Ай бұрын

Thanks for sharing ❤️

Let's use an orthonormal, centerB and first axis (BA). We then have B(0;0) A(c;0) D(c;c) C(0;c) F(a;x) where c is the side length of the square and a = FC. In triangle BCF: BC^2 + FC^2 = BF^2, so c^2 + a^2 = 400. VectorBF(a;c) is orthogonal to (EF) and E is on (EF), so the equation of (EF) is: a.(x -a) + c.(y - c) = 0, or a.x + c.y -a^2 - c^2 =0. Then we have E(c; (a^2 + c^2 -a.c)/c) as E is the point of (EF) whose abscissa is c. Now we have VectorEF(c-a; (a^2 -a.c)/c) We have EF^2 = 144, so (c -a)^2 + ((c - a)^2).((a/c)^2) = 144, or ((c - a)^2).(c^2 + a^2)/c^2) = 144, or ((c - a)^2).(400/c^2) = 144. Then (1 -a/c)^2 = 144/400 = 9/25, and 1 - a/c = 3/5 and a/c = 2/5 and a = (2/5).c. Now, as c^2 + a^2 = 400, we have (c^2) + (4/25).(c^2) = 400, or (c^2).(29/25) = 400, and finally c^2 = (400.25)/29 = 10000/29 and it's the area of the square.

@PreMath

Ай бұрын

Excellent! Thanks for sharing ❤️

Interesting and I got it.

We'll use units of cm and put our units back in at the end BCF and FDE are similar triangles (1/2)*EF*20=120 EF=120/20*2=12 A=LL, L=A^(.5)=BC BF/EF=20/12=5/3=CF/DE=L/DF L=5DF/3 DF=3L/5 LL+FCFC=400 DF+FC=L 3L/5+FC=L FC=2L/5 LL+4LL/25=400 29LL/25=400 LL=10000/29=A=344+24/29 10000(cm)^2/29=(344+24/29)cm^2

Il cateto EF=12.. Posto FC=a..risulta l^2+a^2=400..√(544-l^2)+√(144-(l-a)^2)=l..risulta l^2=89*10^4/2581=344,83..

@PreMath

Ай бұрын

Excellent! Thanks for sharing ❤️

I love Constants ! And K like all Constants of Polynomial Expressions is a Coefficient of X°. 🙂

@PreMath

Ай бұрын

Correct! Thanks for sharing ❤️

The area of square ABCD is A = s². A = (bh)/2 120 = [20(EF)]/2 120 = 10(EF) EF = 12 Label the side length of square ABCD as s. Then, AB = AD = BC = CD = s. Label CF = x. Then, DF = s - x. Let α & β represent the measures of complementary angles. Let m∠CBF = α. Then, because ∠C is a right angle by definition of squares, m∠BFC = β. Because ∠BFE is a right angle, m∠DFE = α. And because ∠D is similarly a right angle, m∠DEF = β. △BCF ~ △FDE by AA. The scale factor of △BCF to △FDE is EF/BF = 12/20 = 3/5. Therefore, DF = 3s/5. Then, s - x = 3s/5. s = 3s/5 + x 5s/5 = 3s/5 + x 2s/5 = x In words, x is 2/5 the length of s. Use the Pythagorean Theorem on △BCF. a² + b² = c² s² + (2s/5)² = 20² s² + (4s²)/25 = 400 25s² + 4s² = 10000 29s² = 10000 s² = 10000/29 (29 is prime, so this complicated fraction cannot be simplified) The area of the blue square is 10000/29 square centimeters (exact), or about 344.83 square centimeters (approximation).

EF＊20/2=120 EF=12 EDF♾️FCB 12 : 20 = 3 : 5 ED=3x DF=3y FC=5x CB=5y 3y+5x=5y 2y=5x y=5x/2 (5x)^2+(25x/2)^2=20^2 25x^2+625x^2/4=400 100x^2+625x^2=1600 725x^2=1600 x^2=64/29 5y=25x/2 area of Blue square : 25x/2＊25x/2=625x^2/4=625/4＊64/29 =10000/29

@PreMath

Ай бұрын

Excellent! Thanks for sharing ❤️

This is an interesting, but not too hard puzzle.😊clearly EF=12, let the square be s×s, then DF=3/5 s, thus FC=2/5 s, ED=3/5×2/5 s=6/25 s, (1+(2/5)^2)s^2=29/25 s^2=20^2, s^2=100^2/29, that is my answer. 😊

@PreMath

Ай бұрын

Very good! Thanks for sharing ❤️

I am not sure about the Solution, because [ABCD] can not be a Square but a Rectangle. The Algebra is correct, but the Geometry denies that [ABCD] is a Square. If one draw a Circle with Center in the Middle Point of EB (Point M), the Diameter will be 12^2 + 20^2 = D^2 ; D^2 = 144 + 400 ; D^2 = 544 ; D^2 = 544 ; D ~ sqrt(544) ; Radius (R) = sqrt(544) / 2 R = sqrt(544/4) ; R = sqrt(136) ; R ~11,66 lin un If I draw a Circle centered in M (Coordinates (0 ; 0)) with Radius sqrt(136) : X^2 + Y^2 = 136, I get 3 Point belonging to the Rectangle [ABCD] : 1) Point B 2) Point E 3) Point F Drawing a Horizontal Line passing through F getting Line DC Drawing a Vertical Line passing through E getting Line AD Drawing two Perpendicular Lines passing through Point B getting Lines AB and BC; BC // AD and AB // DC. If I use this data in a Geometric Calculator I don't get a Square. I get a Rectangle with Sides : AB = CD = 16,492 lin un AD = BC = 19,403 lin un Area [ABCD] = 320 Square Units I may be wrong, but that's the result of my Calculations.

S=10000/29≈344,83

Metric is for people who count on their fingers…

@PreMath

Ай бұрын

Thanks ❤️

Let's find the area: . .. ... .... ..... Since the triangle BEF is a right triangle, we can conclude: A(BEF) = (1/2)*BF*EF ⇒ EF = 2*A(BEF)/BF = 2*(120cm²)/(20cm) = 12cm The triangles BCF and DEF are similar triangles, therefore we obtain: DE/CF = DF/BC = EF/BF = (12cm)/(20cm) = 3/5 The triangles ABE, BCF and DEF are also right triangles, so we can apply the Pythagorean theorem. With s being the side length of the square we obtain: s² + AE² = BE² = BF² + EF² s² + CF² = BF² (s − AE)² + (s − CF²) = EF² s² + AE² = BF² + EF² s² + CF² = BF² s² − 2*s*AE + AE² + s² − 2*s*CF + CF² = EF² −2*s*AE − 2*s*CF = EF² − (BF² + EF²) − BF² = −2*BF² s*(AE + CF) = BF² s*(s − DE + CF) = BF² s*[s − (3/5)*CF + CF] = BF² s*[s + (2/5)*CF] = BF² s² + (2/5)*s*CF = BF² s² + CF² = BF² ∧ s² + (2/5)*s*CF = BF² ⇒ CF = (2/5)*s s² + CF² = BF² s² + (4/25)*s² = BF² (29/25)*s² = BF² ⇒ A(square) = s² = (25/29)*BF² = (25/29)*400cm² = (10000/29)cm² ≈ 344.83cm² Best regards from Germany

@unknownidentity2846

Ай бұрын

Summary: The results is correct, but my solution is slightly too complicated: CF = CD − DF = CD − (3/5)*BC = s − (3/5)*s = (2/5)*s

@PreMath

Ай бұрын

Thanks for sharing ❤️

@JamesDavy2009

Ай бұрын

Equivalency: 1/29 m²

@unknownidentity2846

Ай бұрын

@@JamesDavy2009 That's right, but for those people who want to roughly estimate the side length of the square, I think the decimal value in cm² is the best choice. If you have memorized the square numbers up to 20 (or more), you will immediately see that it must be between 18cm and 19cm. Best regards from Germany

Komennt 1st

@PreMath

Ай бұрын

Excellent! Thanks ❤️